Solving a Quartic Diophantine Equation
Date: 08/20/2005 at 13:09:51 From: Barukh Subject: Quartic Diophantine Equation I would like to know if there is a method of solving the following equation in *rationals*: x^4 + 14*x^2 + 1 = y^2. I tried to re-write the equation as follows: (x^2 + 7)^2 - 48 = y^2, and then solve the equation z^2 - 48 = y^2. There would not be a problem if y and z were integers, but for rationals I don't know how to proceed.
Date: 08/20/2005 at 22:02:12 From: Doctor Vogler Subject: Re: Quartic Diophantine Equation Hi Barukh, Thanks for writing to Dr. Math. That's a very good question. When you have an algebraic equation (a polynomial) in two variables (this is an algebraic curve), and you want to find its rational solutions, this is what you do: First, you determine the genus of the curve. Computing the genus is slightly complicated, and more involved than is usually necessary. So I'll tell you a few special cases and then only go into greater detail if you ask. Every curve has a nonnegative genus, and the largest the genus can be is (d-1)(d-2)/2, where d is the degree of the polynomial. So any polynomial in two variables with degree 1 or 2 has genus zero. Most polynomials of degree three have genus 1 (but not all; the singular ones have genus 0). A curve of the form y^2 = polynomial in x, where the polynomial on the right has no repeated roots, has genus one less than half of the degree of the polynomial in x, rounded up. That's what you have. When you have a curve of genus zero, you can write out a parametric formula for all rational solutions. The most common and the simplest case of this is when you have a degree-two polynomial. To see how to solve this case, refer to this answer from our archives. Rational Solutions to Two Variable Quadratic Equation http://mathforum.org/library/drmath/view/65319.html For other curves of genus zero, refer to this paper by Mark Van Hoeij: http://www.math.fsu.edu/~hoeij/papers/comments/issac1994.html A curve of genus one is called an "elliptic curve." Most of the study of elliptic curves is done by analyzing a degree-three equation of a particular form, called Weierstrass form. Van Hoeij also gives an algorithm to convert any genus-one curve into Weierstrass form: http://www.math.fsu.edu/~hoeij/papers/comments/issac1995.html but for the two most common cases, including yours, I would refer you to the formulas in Connell's manual on elliptic curves, available at http://www.math.mcgill.ca/connell/public/ECH1/ The formulas you need are in chapter 1, section 1.2. For your curve, x^4 + 14*x^2 + 1 = y^2, it says to define the new variables u = (2y + 2)/x^2 v = (4y + 28x^2 + 4)/x^3 because these can give you the x-y values back by the formulas x = (2u + 28)/v y = -1 + (1/2)ux^2, and the equation when converted into u and v becomes a cubic in Weierstrass form. Substitute the formulas for y and then x into your equation, simplify, and you get (as long as x is not zero and u is not -14) v^2 = u^3 + 14u^2 - 4u - 56 = (u - 2)(u + 2)(u + 14). Now you can analyze this elliptic curve. Its Weierstrass parameters are 0, 14, 0, -4, -56 (which will be needed later). It takes a lot of work from the study of elliptic curves to prove that the rational points on an elliptic curve form an abelian group (like you studied in abstract algebra). The law for "adding" points is a geometric line construction described in some detail (with examples) in this answer in our archives: Cubic Diophantine Equation in Three Variables http://mathforum.org/library/drmath/view/66650.html So then the only question that remains is the form of the group, and generators. To make a long story less long, the rational points of any elliptic curve form a group of the form T x Z^r, where T is a finite group (the "torsion" points, all points with finite order) and Z^r is the Cartesian product of r copies of Z (the integers). The number r is called the "rank" of the elliptic curve. Furthermore, there are only a handful of possibilities for the group T, which is either cyclic of rather small order, or is the Cartesian product of a cyclic group of order two with another cyclic group of rather small order. It is actually not very difficult to determine the torsion group of an elliptic curve, but without going into the mathematics of it, I will simply suggest asking the math program GNU Pari, available for download at http://pari.math.u-bordeaux.fr/ You can ask Pari e = ellinit([0,14,0,-4,-56]) to initialize the elliptic curve, and then ask elltors(e) for the torsion group. It will tell you [8, [4, 2], [[10, 48], [-2, 0]]] meaning that the torsion group has order 8, with the form Z/4Z x Z/2Z and with generators (u, v) = (10, 48) and (u, v) = (-2, 0). The other six points in the torsion group can be found by combining these: There is the identity (zero) point, also known as the "point at infinity," and then there are elladd(e,[10,48],[10,48]) elladd(e,[10,48],[2,0]) elladd(e,[10,48],[-2,0]) elladd(e,[2,0],[-2,0]) elladd(e,[10,-48],[-2,0]) which Pari will tell you are [2, 0] [10, -48] [-6, 16] [-14, 0] [-6, -16]. That done, we would like to find out the number r, the rank of the curve. This is, in fact, a much harder problem, and there is (as far as I know) only one program that does this. It is called MWRank, by John Cremona, and can be downloaded from his web page at http://www.maths.nott.ac.uk/personal/jec/ along with a paper describing how it works. You give it 0 14 0 -4 -56 and it gives you a lot more information than you know what to do with, but finally concludes that the rank is zero. That means that the eight torsion points are the only rational points on your curve, namely [10, 48] [-2, 0] [2, 0] [10, -48] [-6, 16] [-14, 0] [-6, -16], and the point at infinity. When we convert these back into x-y coordinates, we get (1, 4), the point at infinity, the point at infinity, (-1, 4), (1, -4), an invalid point, (-1, -4), and another invalid point. It takes some fancy work to show that the two invalid points should actually go to the obvious solutions (0, 1) and (0, -1). So we conclude that these six points are the only solutions to your equation with rational x and y values. It is interesting that the only rational solutions are all integer solutions. In fact, this is common when the rank of the curve is zero, but not when the rank is positive. Given such a simple conclusion, we might want to see if we can prove this in a more elementary fashion. And perhaps we can! Or perhaps not. If you have any questions about this or need more help, please write back and show me what you have been able to do, and I will try to offer further suggestions. - Doctor Vogler, The Math Forum http://mathforum.org/dr.math/
Date: 08/21/2005 at 05:32:48 From: Barukh Subject: Quartic Diophantine Equation Hi Doctor Vogler, Thank you very much for such a detailed explanation. I didn't expect to get such a thorough treatment of the problem. It is just excellent! I was familiar with the elliptic curves method a bit, but didn't know it is suitable for other equations except Weierstrass form. I will definitely look at the links you sent me, and try to perform all the needed steps. If you are curious to know how I arrived at this equation, I can tell you that I tried to find rational solutions to a pair of quadratic DE: r^2 + t^2 = u^2, r^2 + (2t)^2 = v^2. Another QDE that arises from this pair has even simpler form: x^4 - x^2 + 1 = y^2. Best regards, Barukh.
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.